If the first excitation energy of a hydrogen-like atom is 27 . 3 eV , then the ionization energy of this atom will be

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Question: Answered & Verified by Expert

If the first excitation energy of a hydrogen-like atom is

27.3 eV

, then the ionization energy of this atom will be

PhysicsAtomic PhysicsJEE Main

Solution:

1244 Upvotes Verified Answer

The correct answer is: 36.4

First excitation energy

$= Z^{2} Rhc (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = Z^{2} Rhc \frac{3}{4}$

$∴ \frac{3}{4} Rhc = 27.3 eV$

$∴ Z^{2} Rhc = \frac{4}{3} 27.3 eV = 36.4 eV$

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